Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
Answer:
Work done is 0.
Explanation:
Given that,
The circumference of an orbit for a toy on a string is 18 m, r = 18 m
Centripetal force, F = 12 N
In the circular path, the centripetal force is always perpendicular to the motion of the object. Thus it makes an angle of 90 degrees with the force and displacement. Hence, we can say that the centripetal force does not do any work on the toy when it follows its orbit for one cycle.
Answer:
option a is correct
Explanation:
<h2>I hope it's help you ❣️❣️</h2>
Answer:
<h3>The answer is 50 N</h3>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
<h3>force = mass × acceleration</h3>
From the question we have
force = 10 × 5
We have the final answer as
<h3>50 N</h3>
Hope this helps you
